elegant and polished character, for which there is no other outlet. A NOTE ON UNICELLULAR OPERATORS. PETER ROSENTHAL. A bounded linear operator A ...
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A NOTE ON UNICELLULAR OPERATORS PETER ROSENTHAL
A bounded linear operator A on a (complex) Banach space is unicellular it its lattice of invariant subspaces is totally ordered under inclusion. Two examples of unicellular operators are discussed in [l].
Theorem 1. If A is an operator on a Banach space and M is an invariant subspace of A that is comparable with every other invariant subspace of A, then M is invariant under every operator that commutes
with A. Proof. Let AB = BA. Choose a complex number X such that |X| >||7i||. Then (Ti—X)-1 and (73—X) have the same invariant subspaces by a result of Sarason's, [2, p. 53]. The commutativity implies that the subspace (73—X)M is an invariant subspace of A, and thus
that either (73—\)MEMor
iB—X)MZ)M. In the first case the proof
is finished. If (P-X)ilOAf
then AO(73-X)-W,
is invariant
and therefore M
under (73—X)-1 and hence also under (73—X).
Corollary 1. If A is unicellular invariant subspace of A is invariant
and B commutes with A then every under B.
Theorem 2. 7/^4 is a unicellular operator on a separable space then the set of cyclic vectors of A is a residual set.
Banach
Proof. We must show that the set .S of vectors that are not cyclic for A is a first category set. If { Ma} is the family of all proper invariant subspaces of A then clearly S = \\Ma. If the closure of ,S is not the whole space then 5 is contained in a proper subspace and hence is nowhere dense. Suppose, then, that the closure of 5 is the whole space. Let {Xi} be a countable dense subset of S. For each * choose an Mai such that x( is in Mai. It suffices to show that S = [}Mai, for then 5
will be exhibited as a countable
union of nowhere dense sets. Consider
any Ma. If Ma is not contained in \]Mai then Ma contains Mtt. for all i. Thus Ma contains {x»{ and therefore Ma is the whole space.
Corollary 2. If A is a unicellular operator on a separable Banach space then every invariant subspace of A is cyclic. Received by the editors April 14, 1967.
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506
PETER ROSENTHAL
Proof. If M is any invariant subspace of A then A \ M is unicellular and hence Theorem 2 applies. It is easily seen that in the finite-dimensional case an operator is unicellular if and only if it is cyclic and its spectrum contains only one point. An interesting but no doubt difficult question is whether or not every unicellular operator in the infinite-dimensional case must have only one point in its spectrum.
References 1. W. F. Donoghue,
The lattice of invariant subspaces of a completely continuous
quasi-nilpotent transformation, Pacific J. Math. 7 (1957), 1031-1035. 2. Donald Sarason,
The Hp spaces of an annulus,
Mem. Amer. Math. Soc. No. 56
(1965). University of Michigan and University of Toronto
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